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Jim Bryan
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Any three plane conics has a line meeting them in a configuration of points with an order 5 symmetry

Statement: Let $C$ be union of 3 smooth conics in $\mathbb{P}^2$ meeting each other in nodes. Then there exists a line $L\cong \mathbb{P}^1$ such that the configuration of 6 points $x_1,…,x_6\in L$ obtained by intersecting $L$ with $C$ has an automorphism of order 5.

The above fact is surprising (to me at least) since by a dimension count, one cannot get an arbitrary configuration of six points and the configuration of points with an order 5 automorphism is unique. The fact seems to follow from some work a student of mine is doing involving genus 2 K3 surfaces (more below), but the statement seems so classical, that it feels like there should be a direct elementary proof.

Question: Is there a "classical" proof of the above statement?

In fact, the statement seems to hold for any (possibly reducible) sextic curve $C$ with singularities no worse than nodes as long as (1) there is at least one node, and (2) there are no components which are lines.

Background:

Given a sextic plane curve $C$ with at worst nodal singularities, we get a K3 surface by taking the double cover of the plane branched over $C$ and then resolving the ordinary double point singularities. The double covers of the lines in the plane give rise to a two dimensional linear system of genus 2 curves on the K3 which give a rational map $\mathbb{P}^2--\to \overline{M}_2$. One can compute the class of the closure of the image (call it $D$) in the chow ring of $\overline{M}_2$ by intersecting it with test curves in $\overline{M}_2$ (this is tricky so it is possible that we are making mistakes here) and one gets a formula for $D$ as a linear combination of boundary divisors. If $C$ is smooth, the linear combination has integer coefficients, but under the assumption that there is at least one node and no lines in $C$, the coefficients have 5's in the denominator. I think the only way that can happen is if $D$ meets the (unique I think) genus 2 curve with an order 5 automorphism. This gives rise to the statement about the points on the line in $\mathbb{P}^2$.

Question 2: Does the statement hold when $C$ contains lines? For example if $C$ is the union of 6 disjoint lines?

The above proof breaks down in this case.

Jim Bryan
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